- #1

ryoonc

- 5

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## Homework Statement

A horizontal spring with stiffness 0.6 N/m has a relaxed length of 14 cm (0.14 m). A mass of 24 grams (0.024 kg) is attached and you stretch the spring to a total length of 23 cm (0.23 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 14 cm (0.14 m)?

## Homework Equations

Using the Energy Principle:

Kf + Uf = Ki + Ui + W (Kf and W cancel out in this equation), therefore:

(1/2) * ks * sf^2 = [ (1/2) * m * vi^2 ] + [ (1/2) * ks * si^2 ]

## The Attempt at a Solution

I've listed all the variables given:

ks = 0.6 N/m

Lo = 0.14 m

m = 0.024 kg

si = 0.09 m

sf = 0 m

Assuming that I'm looking for the speed of the mass when the spring is at its relaxed length (sf or Lo), I tried plugging in the variables to the energy principle, and got:

0 = [ (1/2) * 0.024kg * Vi^2 ] + [ (1/2) * 0.6 N/m * (0.09 m)^2 ]

However, this confuses me as I thought the question is asking for the final velocity, when the spring is at its relaxed position, and when I try solving this equation for Vi I get a negative number..

Ok so a friend just came by and solved it in 5 minutes, using this equation:

deltaU -> delta K

sqrt[ { (1/2) * 0.6 * (0.09^2) } / m ] = v

Could someone kindly explain the concept to me?

Thanks!